Hydrogen Bonding

8 MCQs9-step worked example
Source: NCERT Chemical Bonding and Molecular StructurePYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

Hydrogen bonding is the single most quietly punishing topic in NEET chemical bonding — not because the concept is hard, but because aspirants underestimate its reach. The question rarely says "hydrogen bond." It says "highest boiling point," "maximum viscosity," or "most soluble in water," and the answer turns on whether you correctly identified where hydrogen bonds form and how strong they are.

What qualifies as a hydrogen bond. A hydrogen bond forms when H is covalently bonded to a small, highly electronegative atom — F, O, or N — and the δ⁺ hydrogen interacts with a lone pair on F, O, or N of a neighbouring molecule. This is an intermolecular (or sometimes intramolecular) electrostatic attraction, not a covalent bond. Typical strength: 10–40 kJ/mol, far stronger than van der Waals forces (~1–5 kJ/mol) but weaker than covalent bonds (~150–400 kJ/mol). NCERT Class 11 Chemistry Chapter 4, page 32 explicitly categorises hydrogen bonding as a special case of dipole-dipole interaction arising from the high electronegativity and small size of F, O, and N.

Two types. Intermolecular hydrogen bonding (e.g., H₂O molecules linking to each other) raises boiling point, viscosity, and surface tension. Intramolecular hydrogen bonding (e.g., the –OH and –NO₂ groups within ortho-nitrophenol) forms a closed ring within one molecule, actually lowering the boiling point relative to para-nitrophenol because it reduces intermolecular association.

The NEET trap pattern. Questions comparing boiling points of HF, H₂O, and NH₃ test whether you know that H₂O has the highest boiling point among the three — not HF, despite F being more electronegative — because each water molecule can form four hydrogen bonds (two via H donors, two via O lone pairs), whereas HF forms only two per molecule in a zig-zag chain. The number of hydrogen bonds per molecule matters more than the electronegativity alone.

Watch out: Cl, S, and C are NOT electronegative or small enough to form classical hydrogen bonds. HCl has no hydrogen bonding — its intermolecular force is dipole-dipole only.

Practice MCQs

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

Which of the following conditions is necessary for the formation of a hydrogen bond?

MCQ 2Easy RecallPractice

Hydrogen bonding is a special case of which type of intermolecular force?

MCQ 3Easy RecallPractice

Which of the following molecules does NOT exhibit hydrogen bonding?

MCQ 4Direct ApplicationPractice

Among HF, H₂O, and NH₃, which has the highest boiling point?

MCQ 5Direct ApplicationPractice

Ortho-nitrophenol has a lower boiling point than para-nitrophenol. The best explanation is:

MCQ 6Direct ApplicationPractice

The typical strength of a hydrogen bond is approximately:

MCQ 7Concept TrapPractice

Ethanol (C₂H₅OH) is miscible with water in all proportions, while dimethyl ether (CH₃OCH₃) has limited water solubility, despite both having the same molecular formula C₂H₆O. The primary reason is:

MCQ 8Concept TrapPractice

Ice floats on water because:

Worked Example

  1. 1

    Given

    Three Group 15 hydrides: PH₃, NH₃, AsH₃.

  2. 2

    Required

    Order of increasing boiling point with reasoning.

  3. 3

    Concept

    Boiling point depends on the strength of intermolecular forces. Among these hydrides, only NH₃ can form hydrogen bonds (H bonded to N, which is small and highly electronegative). PH₃ and AsH₃ have only van der Waals (London dispersion) forces because P and As are neither small enough nor electronegative enough to form hydrogen bonds.

  4. 4

    Framework

    For PH₃ and AsH₃ (no hydrogen bonding): boiling point increases with molecular mass (stronger London dispersion forces). For NH₃: hydrogen bonding provides an additional, much stronger intermolecular attraction that overrides the mass trend.

  5. 5

    Classification of forces

    - PH₃ (M = 34 g/mol): van der Waals only - AsH₃ (M = 78 g/mol): van der Waals only, but higher molecular mass than PH₃ - NH₃ (M = 17 g/mol): van der Waals + hydrogen bonding

  6. 6

    Reasoning

    Between PH₃ and AsH₃, AsH₃ has the larger electron cloud and higher molecular mass, so its London dispersion forces are stronger → higher boiling point. NH₃, despite having the lowest molecular mass of the three, has hydrogen bonding (10–40 kJ/mol per bond) that far outweighs the weak London forces in PH₃ and AsH₃ (~1–5 kJ/mol).

  7. 7

    Final answer

    Increasing boiling point: **PH₃ (−87 °C) < AsH₃ (−55 °C) < NH₃ (−33 °C)**. The expected "mass trend" order would be PH₃ < AsH₃ < NH₃ by mass, but NH₃ is actually the lightest — its anomalously high boiling point is entirely due to hydrogen bonding.

  8. 8

    Common trap

    Aspirants often predict NH₃ should have the lowest boiling point because it has the lowest molecular mass. This fails because hydrogen bonding in NH₃ dominates over the London forces that govern PH₃ and AsH₃.

  9. 9

    Similar NEET-style question

    "Arrange HF, HCl, HBr, HI in order of increasing boiling point." The same logic applies: HCl < HBr < HI follows the mass trend (London forces), but HF is anomalously high due to hydrogen bonding. Answer: HCl < HBr < HI < HF. ---

Before solving, remember these

Key Fact

Hydrogen bonding

Strong dipole-dipole when H is bonded to F, O, or N. Intermolecular (between molecules) or intramolecular (within molecule). Causes high BP of H₂O, HF, NH₃; protein/DNA structure.

-- NCERT Class 11 Chemistry, Ch. 4, p. 32

Formulas

Bond order from MO theory

Higher bond order: shorter, stronger bond. N₂: BO=3, O₂: BO=2, F₂: BO=1.

SymbolQuantitySI Unit
N_bbonding electrons-
N_aantibonding electrons-
BObond order-

Valid when

  • MO theory framework
  • Closed-shell molecule (or with appropriate treatment)

Dipole moment

Product of charge magnitude and bond length. SI: C·m. Common: Debye (1 D = 3.336e-30 C·m).

SymbolQuantitySI Unit
qchargeC
dbond lengthm
mudipole momentC*m or D

Valid when

  • Diatomic or vector-summed for polyatomic
  • Polar bond

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

Trap

Hybridization counting lone pairs

Category: Inorganic Exception

Student counts only bonded atoms when assigning hybridization. Lone pairs count toward steric number too. Steric number = bond pairs + lone pairs → hybridization.

When it triggers

Molecule with central atom having lone pairs (e.g., NH₃: 3 bonds + 1 lp = 4 = sp³; H₂O: 2+2 = 4 = sp³).

How to avoid

Steric number formula: SN = (bond pairs) + (lone pairs). SN=2: sp; SN=3: sp²; SN=4: sp³; SN=5: sp³d; SN=6: sp³d². Lone pairs distort but still count.

Past Year Questions

14 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.

NEET 2025

Given below are two statements : Statement I : A hypothetical diatomic molecule with bond order zero is quite stable. Statement II : As bond order increases, the bond length increases. In the light of the above statements, choose the most appropriate answer from the options given below :

1Statement I is false but Statement II is true
2Both Statement I and Statement II are true
3Both Statement I and Statement II are false
4Statement I is true but Statement II is false
NTA Answer: Option 3(final)
NEET 2025

Given below are two statements : Statement I : Like nitrogen that can form ammonia, arsenic can form arsine. Statement II : Antimony cannot form antimony pentoxide. In the light of the above statements, choose the most appropriate answer from the options given below :

1Statement I is incorrect but Statement II is correct
2Both Statement I and Statement II are correct
3Both Statement I and Statement II are incorrect
4Statement I is correct but Statement II is incorrect
NTA Answer: Option 4(final)
NEET 2022

Which statement regarding polymers is not correct?

1Thermosetting polymers are reusable
2Elastomers have polymer chains held together by weak intermolecular forces
3Fibers possess high tensile strength
4Thermoplastic polymers are capable of repeatedly softening and hardening on heating and cooling respectively
NTA Answer: Option 1(final)

How NEET usually asks this

Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.

Sources

NCERT refs: Class 11 Chemistry Chapter 4, p.32

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